JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
\(50 \,mL\) of \(0.1\, M \,CH _{3} COOH\) is being titrated against \(0.1 \,M\, NaOH\). When \(25\, mL\) of \(NaOH\) has been added, the \(pH\) of the solution will be \(....\,\times 10^{-2}\). (Nearest integer) (Given : \(\left.pK _{ a }\left( CH _{3} COOH \right)=4.76\right)\) \(\log 2=0.30\) \(\log 3=0.48\) \(\log 5=0.69\) \(\log 7=0.84\) \(\log 11=1.04\)
- A \(963\)
- B \(123\)
- C \(476\)
- D \(596\)
Answer & Solution
Correct Answer
(C) \(476\)
Step-by-step Solution
Detailed explanation
Moles of \(CH _{3} COOH =5\, m\, mole\, moles\) of \(NaOH =2.5\, m\, mole\) \(NaOH + CH _{3} COOH \longrightarrow CH _{3} COONa + H _{2} O\) \(2.5\, m\, mole \quad\quad 2.5 \,m\,mole\) \(0 \quad\quad\quad\quad\quad 2.5 m \text { mole } \quad\quad2.5 m \text { mole }\) so buffer…
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